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AleksAgata [21]
3 years ago
10

Given f(x)= 17- x2, what is the average rate of change in f(x) over the interval [1, 5)?

Mathematics
1 answer:
makvit [3.9K]3 years ago
7 0
-6 because it is yeah so
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Given f(x) = -6x - 1 and g(x) = x3, choose<br> the expression for fºg)(x).
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Answer:

- 6x³ - 1

Step-by-step explanation:

Substitute x = g(x) into f(x), that is

(f ○ g)(x)

= f(x³ )

= - 6(x³) - 1

= - 6x³ - 1

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will use 3/4 cup of olive oil to make 3 batches of salad dressing .how much oil does Will use for one batch of salad dressing?
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A boat leaves a port and heads straight East for 12 miles then turns straight south for another 7 miles then stops. If another b
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Step-by-step explanation:

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500 cats were chosen. 52% of the cats liked to sleep inside the house. Fish was the favorite dish of 25% of those cats, while 62
Alex777 [14]
You would need to take 500 * .52 = 260 next take 260 * .25 to find how many cats like fish which equals 65, the probability that it likes fish and it sleeps inside can be found by taking .52 * .25 = .13 so there would be a 13% chance of picking a cat that sleeps indoors and likes fish, 435 cats dont like fish found by taking 500-65=435 and finally to find how many cats sleep outside take 500-260=240 then take 240 * .625 = 150 then 240-150 = 90 so 90 cats like to sleep outside and like fish thats it!=) Hope this helps!
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Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
2 years ago
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